Can the Derivative of a Function be Evaluated as n Approaches Infinity?

[Karlisbad]

Let be a function so the derivative of any order n exist, my question is if there is a way to evaluate:

$$\frac{d^{n}f(x)}{dx^{n}}$$ as $$n\rightarrow \infty$$

 

[Office_Shredder]

Depends on the function. If it's e^x, it's trivial. If it's a polynomial, it's trivial. If it's cos(x), it can't be done (cyclic, so there's no limit). If it's something that's a complete mess of a rational function, you're probably righteously too lazy to figure it out

 

[arildno]

Why should it exist some general way apart from differentiating the function?
The derivatives of the functions f(x)=cos(x) g(x)=1 "evolve" in totally different manners.

 

[Karlisbad]

well i was thinking about "Cauchy's theorem" so you have that the derivative of any order should satisfy:

$$2\pi i f^{(n)}(a)= n! \oint_{C}dzf(z)(z-a)^{-n-1}$$

and from this integral, if C is a circle of unit radius and centered at z=a then, the contour integral just becomes:

$$2\pi f^{(n)}(a)= n! \int_{-\pi}^{\pi} dxf(e^{ix}+a)e^{-inx}$$

EDIT: another good idea would be perhaps to use the "generalized difference" operator so..

$$f^{(n)}(x)=\frac{\nabla ^{n}}{h^{n}}$$ as h-->0 (small h)

 

[arildno]

Why do you assume that Cauchy's theorem in general gives you a neat way to actually evaluate the derivative??

You are not Eljose, are you?

 

[Karlisbad]

I'm supposing that f(z) is analytic with NO poles or at least that the point z=a is not a pole of f, then if f has no poles on the unit circle centered at z=a then "Cauchy Theorem" for derivatives holds.

I don't know what's this stuff about someone called eljose ..and what has to do with me or the forums...